1989 AIME Problems/Problem 14
Problem
Given a positive integer , it can be shown that every complex number of the form
, where
and
are integers, can be uniquely expressed in the base
using the integers
as digits. That is, the equation
![$r+si=a_m(-n+i)^m+a_{m-1}(-n+i)^{m-1}+\cdots +a_1(-n+i)+a_0$](http://latex.artofproblemsolving.com/f/6/4/f6491a5fef34ba26ef6657aae299c94700b40a0e.png)
is true for a unique choice of non-negative integer and digits
chosen from the set
, with
. We write
![$r+si=(a_ma_{m-1}\ldots a_1a_0)_{-n+i}$](http://latex.artofproblemsolving.com/5/6/a/56aa9ca7c641491fbd454fff558b7a8fb8918ced.png)
to denote the base expansion of
. There are only finitely many integers
that have four-digit expansions
![$k=(a_3a_2a_1a_0)_{-3+i}~~$](http://latex.artofproblemsolving.com/f/8/1/f81d425eb93665b6c8598c7cf275730089c4df4d.png)
![$~~a_3\ne 0.$](http://latex.artofproblemsolving.com/0/6/3/063a24c5b1954678bd93bbec326e554e671a4acf.png)
Find the sum of all such ,
Solution
First, we find the first three powers of :
So we solve the diophantine equation .
The minimum the left-hand side can go is -54, so since
can't equal 0, so we try cases:
- Case 1:
- The only solution to that is
.
- Case 2:
- The only solution to that is
.
So we hafour-digitr digit integers and
, and we need to find the sum of all integers
that can be expressed by one of those.
:
We plug the first three digits into base 10 to get . The sum of the integers
in that form is
.
:
We plug the first three digits into base 10 to get . The sum of the integers
in that form is
. The answer is
.
~minor edit by Yiyj1
See also
1989 AIME (Problems • Answer Key • Resources) | ||
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Followed by Problem 15 | |
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